# Solve the Differential Equation #y'' - 4 y' + 4y = 2 e^(2x)#?

Method of Undetermined Coefficients

Start with the homogeneous equation and the complementary solution :

This has characteristic equation:

Putting these into the equation:

Now applying the IV's:

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The solution is

This is a second order linear, non-homogeneous differential equation.

The general solution can be written as

The caracteristic equation is

We have a double root

The solution without the LHS is

For the particular solution, we try

Putting these in the equation

So, the particular solution is

The general solution is

Finally, we have

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The two other solutions quite clearly demonstrate how to solve the complementary solution of the homogeneous equation.

As this is fairly standard text book stuff which solves the Auxiliary equation to form a guaranteed solution based of the roots of the equation

Basically it is down to practice & experience but there is a solid method to find the particular solution using the Wronskian. It does, however, involve a lot more work:

Having established that the solution of the homogeneous equation is (see other answers):

is given by:

Where:

So for our equation:

So the wronskian for this equation is:

So we form the two particular solution function:

And;

And so we form the Particular solution:

Which is the same particular solution as the other answers produced, leading to the general solution:

Which then leads to the same specific solution of the other answers.

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Laplace Transform

From Standard Tables, here using Mary Boas 3rd Ed, L6 and L18, pp469-470:

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Another approach using the fact that

Mary Boas teaches this in a way I not see very widely used, but which also obviates the need for the experienced guesswork.

And then apply the IV's as before.

Boas is my personal favourite maths book, but it's for scientists, not mathematicians. But this is a real good way to go at repeated eigenvalues, which are tricky.

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To solve the differential equation (y'' - 4y' + 4y = 2e^{2x}), we can use the method of undetermined coefficients. The complementary solution is (y_c = c_1e^{2x} + c_2xe^{2x}) since the characteristic equation (r^2 - 4r + 4 = 0) has a repeated root at (r = 2).

For the particular solution, we assume (y_p = Ae^{2x}) because (2e^{2x}) is similar to the complementary solution term (e^{2x}), and we multiply it by a constant (A) to avoid duplication.

Differentiating (y_p) twice and substituting back into the original equation, we get:

[y''_p - 4y'_p + 4y_p = 2e^{2x}] [4Ae^{2x} - 8Ae^{2x} + 4Ae^{2x} = 2e^{2x}] [0 = 2e^{2x}]

This is not possible, so we need to include another term. Let's assume (y_p = Bxe^{2x}).

Differentiating (y_p) twice and substituting back into the original equation, we get:

[y''_p - 4y'_p + 4y_p = 2e^{2x}] [4Be^{2x} + 4Bxe^{2x} - 8Be^{2x} - 8Bxe^{2x} + 4Bxe^{2x} = 2e^{2x}] [4Be^{2x} = 2e^{2x}]

[4B = 2]

[B = \frac{1}{2}]

So, the particular solution is (y_p = \frac{1}{2}xe^{2x}).

The general solution is the sum of the complementary and particular solutions:

[y = y_c + y_p] [y = c_1e^{2x} + c_2xe^{2x} + \frac{1}{2}xe^{2x}]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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